Problem 88 - Calculus and its applications

Question

Differentiate. $$ y=\left(\frac{x}{\sqrt{x-1}}\right)^{3} $$

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{dx} = \frac{3x^2(x-2)}{2(x-1)^3} \).

Step 1: Apply the Chain Rule

To differentiate the function \( y = \left(\frac{x}{\sqrt{x-1}}\right)^3 \), we begin by applying the chain rule. Let \( u = \frac{x}{\sqrt{x-1}} \), so \( y = u^3 \). Differentiating with respect to \( x \) gives \( \frac{dy}{dx} = 3u^2 \cdot \frac{du}{dx} \).

Step 2: Differentiate the Inner Function u

Next, differentiate \( u = \frac{x}{\sqrt{x-1}} \) with respect to \( x \). Use the quotient rule: \( \frac{du}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \), where \( u = x \) and \( v = \sqrt{x-1} \). Then \( \frac{du}{dx} = 1 \) and \( \frac{dv}{dx} = \frac{1}{2\sqrt{x-1}} \).

Step 3: Compute the Derivative of u

Insert the derivatives computed into the quotient rule: \[\frac{du}{dx} = \frac{\sqrt{x-1} \cdot 1 - x \cdot \frac{1}{2\sqrt{x-1}}}{(\sqrt{x-1})^2} = \frac{\sqrt{x-1} - \frac{x}{2\sqrt{x-1}}}{x-1}. \] Simplify this expression to find \( \frac{du}{dx} \).

Step 4: Simplify du/dx

To simplify \( \frac{du}{dx} \), multiply the entire numerator and denominator by \( 2\sqrt{x-1} \) to remove the square root:\[\frac{du}{dx} = \frac{2(x-1) - x}{2(x-1)^{3/2}} = \frac{2x - 2 - x}{2(x-1)^{3/2}} = \frac{x - 2}{2(x-1)^{3/2}}. \]

Step 5: Compute dy/dx

Now insert \( \frac{du}{dx} \) back into the chain rule equation: \[\frac{dy}{dx} = 3u^2 \cdot \frac{x - 2}{2(x-1)^{3/2}}. \] Substitute \( u = \frac{x}{\sqrt{x-1}} \) back in:\[\frac{dy}{dx} = 3 \left(\frac{x}{\sqrt{x-1}}\right)^2 \cdot \frac{x - 2}{2(x-1)^{3/2}}= \frac{3x^2(x-2)}{2(x-1)^3}. \]

Step 6: Simplify the Final Expression

The final expression for \( \frac{dy}{dx} \) is simplified as:\[\frac{dy}{dx} = \frac{3x^{2}(x-2)}{2(x-1)^3}. \] This is the derivative of the original function in its simplest form.

Key Concepts

Chain Rule

The Chain Rule is a fundamental tool in calculus used to differentiate composite functions. If you have a function that is itself a function of another variable, the Chain Rule helps you find its derivative. In simple terms, if you have a function \( y = f(g(x)) \), then the derivative with respect to \( x \) is given by:\[ \frac{dy}{dx} = f'(g(x)) \, g'(x) \]This rule is crucial when dealing with nested functions, like \( y = \left(\frac{x}{\sqrt{x-1}}\right)^3 \). Here, it was helpful to set the inner function \( u = \frac{x}{\sqrt{x-1}} \) and treat \( y = u^3 \). By applying the Chain Rule, the differentiation started from the outside function and carefully moved inward, breaking the process into more manageable steps.

Quotient Rule

The Quotient Rule is essential when you divide two functions and need to find the derivative of their quotient. If you have two functions represented as \( u(x) \) and \( v(x) \), their quotient is \( \frac{u}{v} \). The rule for differentiation is:\[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]In the given expression \( u = \frac{x}{\sqrt{x-1}} \), the numerator and the denominator both require differentiation. To apply the Quotient Rule:- \( u = x \) and \( v = \sqrt{x-1} \)- \( \frac{du}{dx} = 1 \)- \( \frac{dv}{dx} = \frac{1}{2\sqrt{x-1}} \)These derivatives are then plugged into the Quotient Rule formula to systematically find the derivative of \( u \). This step-by-step approach helps avoid confusion and delivers a clear path to the simplified derivative.

Function Derivatives

Understanding function derivatives is a cornerstone of calculus. Derivatives measure how a function changes as its input changes. Differentiation is about finding this rate of change or the "instantaneous rate of change" along the function. When dealing with complex functions, separate them into simpler parts. Differentiate each part individually, using rules like the Chain Rule or Quotient Rule.
For instance, when differentiating \( y = \left(\frac{x}{\sqrt{x-1}}\right)^3 \), we used both rules:

• Applied the Chain Rule first, focusing on the outer function \( y = u^3 \)
• Applied the Quotient Rule for the inner function \( u = \frac{x}{\sqrt{x-1}} \)

Combining these methods illustrates the elegance of function differentiation and provides tools to handle increasingly intricate derivatives.

Simplification of Expressions

After performing differentiation, the resulting expressions often need simplification. Simplification makes derivatives easier to interpret and use. It involves reducing terms, canceling out units, and ensuring that the expression is in its most compact form.
When solving \( \frac{dy}{dx} \) for the function \( y = \left(\frac{x}{\sqrt{x-1}}\right)^3 \), simplification required strategic manipulation:

• Combining like terms and removing unnecessary complex fractions
• Factoring out common elements, such as powers and coefficients
• Resulting in an expression like \( \frac{3x^2(x-2)}{2(x-1)^3} \)

This final derivative is both simpler and more insightful, aiding in further analysis or application. Effective simplification sharpens our mathematical insight and provides clarity in the presentation of results.